Re: proof that inverse of continuous function is also continuous?

On Tue, 20 Dec 2005 20:28:40 -0500, "Stephen J. Herschkorn"
<sjherschko@xxxxxxxxxxxx> wrote:

>David C. Ullrich wrote:
>
>>On 19 Dec 2005 15:04:50 -0800, i.love.jeevitha@xxxxxxxxx wrote:
>>
>>
>>
>>>Does anyone know a non-analysis proof? e-d proofs are okay, without
>>>the mention of metric spaces and such.
>>>
>>>
>>
>>The fact you want to prove is _false_ in many situations,
>>so a proof using metric spaces or such is not likely to exist.
>>
>>It's true if we're talking about functions from R to R,
>>or from an interval to R. In that case (as suggested)
>>you can use the fact that if f has an inverse then f
>>must be monotone (either increasing or decreasing).
>>
>
>And what is a "non-analysis" proof of the fact that a continuous
>injection from R to R is strictly monotone?

I didn't claim to be giving a "non-analysis" proof of anything.
It's been pointed out that the phrase doesn't really make
much sense - the OP finally said that all he(?) meant was
a proof using nothing execpt what high-school students
usually call calculus. Which of course is impossible,
since there are no proofs at all in what high-school
students usually call calculus.

I'm not sure what the real point to your question was,
so I'm not sure whether the following is what you wanted
to know or something your already know every well:

It _is_ possible to give a simple proof of this fact
based on something that the typical high-school
calculus student will agree is clearly true, namely
the Intermediate Value Theorem:

Suppose that f : R -> R is injective but not monotone.
Then there exist a < b with f(a) < f(b) and there
exist c < d with f(c) > f(d).

This, plus the IVT, contradicts the fact that f is 1-1.
I'm not familiar with too many books at this level;
I know of one book that gets the contradiction by
breaking things into a large number of cases
depending on various inequalities involving
a, ... b and f(a), ... f(b). One can get the
contradiction from a "trick" one-case argument,
I don't know whether I invented it or not:

Define

g(t) = f(ta + (1-t)c) - f(tb + (1-t)d).

Then g(0) > 0 and g(1) < 0, so there exists
a t in (0,1) such that g(t) = 0. The fact
that g(t) = 0 (and 0 < t < 1) shows that f
is not 1-1.


************************

David C. Ullrich
.

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